选择不仅关系到另一个注册表

Question

我模仿一个小型的数据库更容易解释最后插入的项目：

CREATE TABLE bands (
    id INTEGER UNSIGNED NOT NULL AUTO_INCREMENT, 
    name VARCHAR(120) NULL, 
    PRIMARY KEY(id) 
) 
TYPE=InnoDB; 

CREATE TABLE albums (
    id INTEGER UNSIGNED NOT NULL AUTO_INCREMENT, 
    band_id INTEGER UNSIGNED NOT NULL, 
    album_name VARCHAR(120) NULL, 
    rating INTEGER UNSIGNED NULL, 
    insertion_date TIMESTAMP NULL, 
    PRIMARY KEY(id), 
    INDEX albums_FKIndex1(band_id), 
    FOREIGN KEY(band_id) 
    REFERENCES bands(id) 
     ON DELETE NO ACTION 
     ON UPDATE NO ACTION 
) 
TYPE=InnoDB; 

现在，假装我们已经有一些乐队和他们各自的表中登记的多张唱片，我想仅选择每个已注册乐队中最后插入的专辑。

PS：我必须使用“album.insertion_date”字段来确定哪个相册是最后插入的。

Answer 1

尝试通过insertion_date和带连接两个表和筛选：

SELECT al.* 
FROM albums al 
INNER JOIN bands b ON al.band_id=b.id 
WHERE al.insertion_date=(
    SELECT max(insertion_date) 
    FROM albums 
    WHERE band_id=b.id 
) 

Answer 2

试试这个：

select b.name, a.album_name, a.isertion_date 
    from bands b, albums a 
    where a.band_id = b.id 
    and a.insertion_date = (select max(a1.insertion_date) from albums a1 where a1.band_id = b.id) 

Answer 3

考虑到你的相册IDS是AUTO_INCREMENT和可能性为insertion_date是NULL（因为它是默认值），使用insertion_date判定结果不这样做，但最聪明的事...你去：

SELECT DISTINCT band, last_album, insertion_date 
FROM (
     SELECT bands.name AS band, albums.album_name AS last_album, albums.insertion_date 
     FROM bands 
     JOIN albums ON bands.id=albums.band_id 
     ORDER BY albums.insertion_date DESC 
    ) t1 
GROUP BY band;